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anonymous155 · Sep 5, 2012, 06:04 AM

(1+cosx / 1-cosx) - (1-cosx/1+cosx) = (4cscx)(cotx)

ebaines · Sep 5, 2012, 06:13 AM

For clarification with proper use of parentheses the identity is this:

(1+cosx)/(1-cosx) - (1-cosx)/(1+cosx) = 4 cscx cotx

Showing this is simply a matter of combining the two fractions and simplifying. Show us your attempt and we'll help you along.

anonymous155 · Sep 5, 2012, 06:28 AM

Originally Posted by ebaines

For clarification with proper use of parentheses the identity is this:

(1+cosx)/(1-cosx) - (1-cosx)/(1+cosx) = 4 cscx cotx

Showing this is simply a matter of combining the two fractions and simplifying. Show us your attempt and we'll help you along.

Ok so combining these two fractions on the left the two cosines would cancel out right? Or should I convert 1-cos x into sin^2 x? And then I would change csc x to 1/sin x. Making it 4/sin x. Change cot x to 1/tan x then cross multiply to get 4tan x /sin x. Tell me if Im on the right track

ebaines · Sep 5, 2012, 08:38 AM

Cosines do NOT cancel out, and 1-cosx doed not equal sin^2x.

To combine these fractions multiply the first one by (1+c)/(1+c) and the second one by (1-c)(1-c), and that way they end up with common denominators:

$<br /> \frac {(1+\cos x)}{(1-\cos x)} + \frac {(1- \cos x)}{(1 + \cos x)} = \frac {(1+\cos x)}{(1-\cos x)} \times \frac {(1 + \cos x)}{(1+ \cos x) } + \frac {(1- \cos x)}{(1 + \cos x)} \times \frac {(1- \cos x)}{(1 - \cos x)} = \frac {(1+\cos x)^2 - (1- \cos x)^2}{(1 - \cos x)(1 + \cos x)}<br />$

Now continue to simplify and see where it takes you.